Question

German physicist Werner Heisenberg related the uncertainty of an object's position (Δx) to the uncertainty in its velocity (Δ????) Δx≥ℎ4????mΔ???? where ℎ is Planck's constant and m is the mass of the object. The mass of an electron is 9.11×10−31 kg. What is the uncertainty in the position of an electron moving at 4.00×106 m/s with an uncertainty of Δ????=0.01×106 m/s?

Answer 1

Answer:

$\Delta x = 5.47 \times 10^{-9} m$

Explanation:

As we know by the principle of uncertainty that the product of uncertainty in position and uncertainty in momentum is given as

$\Delta x \times \Delta P = \frac{h}{4\pi}$

so here we know that

$\Delta v = 0.01 \times 10^6 m/s$

$m = 9.11 \times 10^{-31} kg$

so we have

$\Delta x \times (9.11 \times 10^{-31})(0.01 \times 10^6) = \frac{6.26 \times 10^{-34}}{4\pi}$

$\Delta x = 5.47 \times 10^{-9} m$